Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer:
<em>»</em><em> </em><em>λ </em><em>=</em><em> </em><em>0</em><em>.</em><em>0</em><em>2</em><em>m</em>
Explanation:
Given :
Velocity of the wave {v}= 12 m/s
Frequency {f} = 600 Hz
Apply Wavelength formula :
• 
→ λ = 
→ λ = 
→ λ = 0.02m
Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
Answer:
The concentration of this sodiumhydroxide solutions is 0.50 M
Explanation:
Step 1: Data given
Mass of sodium hydroxide (NaOh) = 8.0 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Volume water = 400 mL = 0.400 L
Step 2: Calculate moles NaOH
Moles NaOH = mass NaOH / molar mass NaOH
Moles NaOH = 8.0 grams / 40.0 g/mol
Moles NaOh = 0.20 moles
Step 3: Calculate concentration of the solution
Concentration solution = moles NaOH / volume water
Concentration solution = 0.20 moles / 0.400 L
Concentration solution = 0.50 M
The concentration of this sodiumhydroxide solutions is 0.50 M
